Write $3n+1=x^2$ Then $(n+1)=\frac{(x-1)^2}{9}+\frac{(x-1)^2}{9}+\frac{(x+2)^2}{9}$

$3n+1 = x^2 \implies x^2 \equiv 1 \pmod 3 \implies x = \pm 1 \pmod 3 = 3k \pm 1$ for some $k \in \mathbb N$. $\implies 3n + 1 = 9k^2 \pm 6k + 1 \implies n = 3k^2 \pm 2k \implies n + 1 = 3k^2 \pm 2k + 1 = (k \pm 1)^2 + k^2 + k^2$.